Ganit Prakash - Class-X - Compound Interest
Let us work out 6.1 Uniform Rate of Increase or Decrease
📘 Exercise 6.1 Solutions
Compound InterestGiven:
Principal \((P)\) = ₹ 5000
Rate of compound interest \((r)\) = 8.5% per annum
Time \((t)\) = 2 years
Amount at the end of 2 years \((A)\) = \( P\left(1 + \frac{r}{100}\right)^t \)
\[\begin{array}{l} A = 5000 \left(1 + \frac{8.5}{100}\right)^2 \\ = 5000 \left(1 + \frac{85}{1000}\right)^2 \\ = 5000 \left(\frac{1085}{1000}\right)^2 \\ = 5000 \times \frac{1085 \times 1085}{1000 \times 1000} \\ = \frac{5886125}{1000} \\ = 5886.125 \end{array}\]Therefore, I shall get ₹ 5886.13 (approx) at the end of 2 years.
Given:
Principal \((P)\) = ₹ 5000
Rate of compound interest \((r)\) = 8% per annum
Time \((t)\) = 3 years
Amount at the end of 3 years \((A)\) = \( P\left(1 + \frac{r}{100}\right)^t \)
\[\begin{array}{l} A = 5000 \left(1 + \frac{8}{100}\right)^3 \\ = 5000 \left(\frac{100+8}{100}\right)^3 \\ = 5000 \left(\frac{108}{100}\right)^3 \\ = 5000 \times \frac{108 \times 108 \times 108}{100 \times 100 \times 100} \\ = \frac{27 \times 108 \times 54}{25} \\ = \frac{157464}{25} \\ = 6298.56 \end{array}\]Therefore, the required amount is ₹ 6298.56.
Given:
Principal \((P)\) = ₹ 2000
Rate of compound interest \((r)\) = 6% per annum
Time \((t)\) = 2 years
First, let us calculate the Amount \((A)\) at the end of 2 years:
\[\begin{array}{l} A = 2000 \left(1 + \frac{6}{100}\right)^2 \\ = 2000 \left(\frac{100+6}{100}\right)^2 \\ = 2000 \left(\frac{106}{100}\right)^2 \\ = 2000 \times \frac{106 \times 106}{100 \times 100} \\ = 2247.2 \end{array}\]Compound Interest \((CI)\) = Amount - Principal
\[\begin{array}{l} CI = 2247.2 - 2000 \\ = 247.20 \end{array}\]Therefore, Goutam babu will pay a compound interest of ₹ 247.20.
Given:
Principal \((P)\) = ₹ 30000
Rate of compound interest \((r)\) = 9% per annum
Time \((t)\) = 3 years
First, calculate the Amount \((A)\):
\[\begin{array}{l} A = P\left(1 + \frac{r}{100}\right)^t \\ = 30000 \left(1 + \frac{9}{100}\right)^3 \\ = 30000 \left(\frac{109}{100}\right)^3 \\ = 30000 \times \frac{109 \times 109 \times 109}{100 \times 100 \times 100} \\ = 3 \times \frac{1295029}{10} \\ = 38850.87 \end{array}\]Now, calculate Compound Interest \((CI)\):
\[\begin{array}{l} CI = Amount - Principal \\ CI = 38850.87 - 30000 \\ = 8850.87 \end{array}\]Therefore, the compound interest is ₹ 8850.87.
Given:
Principal \((P)\) = ₹ 80000
Rate of compound interest \((r)\) = 5% per annum
Time \((t)\) = \(2\frac{1}{2}\) years
For fractional years, the Amount \((A)\) is calculated as:
\( A = P \left(1 + \frac{r}{100}\right)^n \times \left(1 + \frac{\frac{f}{m}r}{100}\right) \)
Here, whole years \((n)\) = 2, and fraction of year \((\frac{f}{m})\) = \(\frac{1}{2}\).
\[\begin{array}{l} A = 80000 \left(1 + \frac{5}{100}\right)^2 \times \left(1 + \frac{2.5}{100}\right) \\ = 80000 \left(\frac{105}{100}\right)^2 \times \left(\frac{102.5}{100}\right) \\ = 80000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{1025}{1000} \\ = 80000 \times \frac{441}{400} \times \frac{41}{40} \\ = 200 \times 441 \times \frac{41}{40} \\ = 5 \times 441 \times 41 \\ = 2205 \times 41 \\ = 90405 \end{array}\]Therefore, the required amount is ₹ 90405.
Given:
Let the principal \((P)\) be ₹ \(x\).
Rate of compound interest \((r)\) = 8% per annum
Time \((t)\) = 2 years
Compound Interest \((CI)\) = ₹ 2496
We know the formula for Compound Interest is:
\( CI = P \left[ \left(1 + \frac{r}{100}\right)^t - 1 \right] \)
\[\begin{array}{l} 2496 = P \left[ \left(1 + \frac{8}{100}\right)^2 - 1 \right] \\ 2496 = P \left[ \left(\frac{108}{100}\right)^2 - 1 \right] \\ 2496 = P \left[ \left(\frac{27}{25}\right)^2 - 1 \right] \\ 2496 = P \left[ \frac{729}{625} - 1 \right] \\ 2496 = P \left( \frac{729 - 625}{625} \right) \\ 2496 = P \times \frac{104}{625} \\ P = \frac{2496 \times 625}{104} \\ P = 24 \times 625 \\ P = 15000 \end{array}\]Therefore, Chandadavi had borrowed ₹ 15000.
Let the required principal = ₹ \(x\)
Therefore, amount of ₹ \(x\) for 3 years at 10% compound interest per annum
\[\begin{array}{l} = x \times \left(1 + \frac{10}{100}\right)^3 \\ = x \times \left(\frac{100+10}{100}\right)^3 \\ = x \times \left(\frac{110}{100}\right)^3 \\ = x \times \left(\frac{11}{10}\right)^3 \\ = x \times \frac{1331}{1000} = \frac{1331x}{1000} \end{array}\]Therefore, compound interest,
\[\begin{array}{l} \left(\frac{1331x}{1000} - x\right) \\ = \left(\frac{1331x - 1000x}{1000}\right) \\ = \frac{331x}{1000} \end{array}\]As per question, \(\frac{331x}{1000} = 2648\)
\[\begin{array}{l} \Rightarrow x = \frac{2648 \times 1000}{331} \\ \Rightarrow x = 8000 \end{array}\]Therefore, the required principal is ₹ 8000.
Let Rahaman chacha had deposited ₹ \(x\) in the bank.
Therefore, amount of ₹ \(x\) for 2 years at 9% compound interest per annum
\[\begin{array}{l} = x \times \left(1 + \frac{9}{100}\right)^2 \\ = x \times \left(\frac{100+9}{100}\right)^2 \\ = x \times \left(\frac{109}{100}\right)^2 \end{array}\]As per question, \(x \times \left(\frac{109}{100}\right)^2 = 29702.50\)
\[\begin{array}{l} \Rightarrow x \times \frac{109 \times 109}{100 \times 100} = \frac{2970250}{100} \\ \Rightarrow x = \frac{2970250 \times 100}{109 \times 109} \\ \Rightarrow x = 25000 \end{array}\]Therefore, he had deposited ₹ 25000 in the bank.
Let the required principal = ₹ \(x\)
Therefore, amount of ₹ \(x\) for 3 years at 8% compound interest per annum
\[\begin{array}{l} = x \times \left(1 + \frac{8}{100}\right)^3 \\ = x \times \left(\frac{100+8}{100}\right)^3 \\ = x \times \left(\frac{108}{100}\right)^3 \end{array}\]As per question, \(x \times \left(\frac{108}{100}\right)^3 = 31492.80\)
\[\begin{array}{l} \Rightarrow x \times \frac{108 \times 108 \times 108}{100 \times 100 \times 100} = \frac{3149280}{100} \\ \Rightarrow x = \frac{3149280 \times 100 \times 100}{108 \times 108 \times 108} \\ \Rightarrow x = 25000 \end{array}\]Therefore, the required principal is ₹ 25000.
Principal \((p)\) = ₹ 12000
Rate of interest \((r\%)\) = 7.5%
Time \((t)\) = 2 years
Compound interest of ₹ 12000 for 2 years at 7.5% per annum
\[\begin{array}{l} = 12000 \left\{ \left(1 + \frac{7.5}{100}\right)^2 - 1 \right\} \\ = 12000 \left\{ \left(\frac{1075}{1000}\right)^2 - 1 \right\} \\ = 12000 \left\{ \left(\frac{43}{40}\right)^2 - 1 \right\} \\ = 12000 \left\{ \frac{1849 - 1600}{1600} \right\} \\ = 12000 \times \frac{249}{1600} = \frac{3735}{2} = 1867.50 \end{array}\]Simple interest of ₹ 12000 for 2 years at 7.5% per annum \(= \frac{P \times r \times t}{100}\)
\[\begin{array}{l} = \left(12000 \times \frac{7.5}{100} \times 2\right) = 1800 \end{array}\]Therefore, difference between compound interest and simple interest \(= (1867.50 - 1800) = 67.50\)
Principal \((p)\) = ₹ 10000
Rate of interest \((r\%)\) = 5%
Time \((t)\) = 3 years
Compound interest of ₹ 10000 for 3 years at 5% per annum
\[\begin{array}{l} = 10000 \left\{ \left(1 + \frac{5}{100}\right)^3 - 1 \right\} \\ = 10000 \left\{ \left(\frac{105}{100}\right)^3 - 1 \right\} \\ = 10000 \left\{ \left(\frac{21}{20}\right)^3 - 1 \right\} \\ = 10000 \left\{ \frac{9261 - 8000}{8000} \right\} \\ = 10000 \times \frac{1261}{8000} = \frac{12610}{8} = 1576.25 \end{array}\]Simple interest of ₹ 10000 for 3 years at 5% per annum \(= \frac{P \times r \times t}{100}\)
\[\begin{array}{l} = \left(10000 \times \frac{5}{100} \times 3\right) = 1500 \end{array}\]Therefore, difference between compound interest and simple interest \(= (1576.25 - 1500) = 76.25\)
Let the required sum of money = ₹ \(x\)
Rate of interest \((r\%)\) = 9%
Time \((t)\) = 2 years
Compound interest of ₹ \(x\) for 2 years at 9% per annum
\[\begin{array}{l} = x \left\{ \left(1 + \frac{9}{100}\right)^2 - 1 \right\} \\ = x \left\{ \left(\frac{109}{100}\right)^2 - 1 \right\} \\ = x \left\{ \frac{11881 - 10000}{10000} \right\} \\ = x \times \frac{1881}{10000} = \frac{1881x}{10000} \end{array}\]Simple interest of ₹ \(x\) for 2 years at 9% per annum \(= \frac{x \times 9 \times 2}{100} = \frac{18x}{100} = \frac{1800x}{10000}\)
Therefore, difference between compound interest and simple interest,
\[\begin{array}{l} = \left(\frac{1881x}{10000} - \frac{1800x}{10000}\right) \\ = \frac{81x}{10000} \end{array}\]As per question, \(\frac{81x}{10000} = 129.60\)
\[\begin{array}{l} \Rightarrow 81x = 129.60 \times 10000 \\ \Rightarrow x = \frac{1296000}{81} \\ \Rightarrow x = 16000 \end{array}\]Therefore, the required sum of money is ₹ 16000.
Let the required sum of money = ₹ \(x\)
Rate of interest \((r\%)\) = 10%
Time \((t)\) = 3 years
Compound interest of ₹ \(x\) for 3 years at 10% per annum
\[\begin{array}{l} = x \left\{ \left(1 + \frac{10}{100}\right)^3 - 1 \right\} \\ = x \left\{ \left(\frac{110}{100}\right)^3 - 1 \right\} \\ = x \left\{ \left(\frac{11}{10}\right)^3 - 1 \right\} \\ = x \left\{ \frac{1331 - 1000}{1000} \right\} = \frac{331x}{1000} \end{array}\]Simple interest of ₹ \(x\) for 3 years at 10% per annum \(= \frac{x \times 10 \times 3}{100} = \frac{30x}{100} = \frac{300x}{1000}\)
Therefore, difference between compound interest and simple interest,
\[\begin{array}{l} = \left(\frac{331x}{1000} - \frac{300x}{1000}\right) \\ = \frac{31x}{1000} \end{array}\]As per question, \(\frac{31x}{1000} = 930\)
\[\begin{array}{l} \Rightarrow 31x = 930 \times 1000 \\ \Rightarrow x = \frac{930000}{31} \\ \Rightarrow x = 30000 \end{array}\]Therefore, the required sum of money is ₹ 30000.
Principal \((P)\) = ₹ 6000
Rate of interest for 1st year \((r_1)\) = 7%
Rate of interest for 2nd year \((r_2)\) = 8%
Therefore, amount of ₹ 6000 after 2 years
\[\begin{array}{l} = 6000 \times \left(1 + \frac{7}{100}\right) \left(1 + \frac{8}{100}\right) \\ = 6000 \times \left(\frac{107}{100}\right) \times \left(\frac{108}{100}\right) \\ = 6000 \times \frac{107 \times 108}{10000} \\ = \frac{6 \times 11556}{10} \\ = \frac{69336}{10} = 6933.60 \end{array}\]Therefore, Compound Interest = Amount - Principal
\[\begin{array}{l} = (6933.60 - 6000) \\ = 933.60 \end{array}\]Therefore, the required compound interest is ₹ 933.60.
Principal \((P)\) = ₹ 5000
Rate of interest for 1st year \((r_1)\) = 5%
Rate of interest for 2nd year \((r_2)\) = 6%
Therefore, amount of ₹ 5000 after 2 years
\[\begin{array}{l} = 5000 \times \left(1 + \frac{5}{100}\right) \left(1 + \frac{6}{100}\right) \\ = 5000 \times \left(\frac{105}{100}\right) \times \left(\frac{106}{100}\right) \\ = 5000 \times \frac{21}{20} \times \frac{53}{50} \\ = \frac{5000 \times 1113}{1000} \\ = 5 \times 1113 = 5565 \end{array}\]Therefore, Compound Interest = Amount - Principal
\[\begin{array}{l} = (5565 - 5000) \\ = 565 \end{array}\]Therefore, the required compound interest is ₹ 565.
Let the required sum of money = ₹ \(x\) and rate of interest = \(r\%\) per annum.
\[\begin{array}{l} \text{Simple interest for 1 year } = \frac{x \times r \times 1}{100} = \frac{xr}{100} \\ \text{As per question, } \frac{xr}{100} = 50 \Rightarrow xr = 5000 \dots (i) \end{array}\] \[\begin{array}{l} \text{Compound interest for 2 years } = x \left\{ \left(1 + \frac{r}{100}\right)^2 - 1 \right\} \\ = x \left\{ 1 + \frac{2r}{100} + \frac{r^2}{10000} - 1 \right\} \\ = x \left( \frac{2r}{100} + \frac{r^2}{10000} \right) \\ = \frac{2xr}{100} + \frac{xr \cdot r}{10000} \end{array}\] \[\begin{array}{l} \text{As per question, } \frac{2xr}{100} + \frac{xr \cdot r}{10000} = 102 \\ \Rightarrow \frac{2 \times 5000}{100} + \frac{5000 \times r}{10000} = 102 \quad [\text{From (i)}] \\ \Rightarrow 100 + \frac{r}{2} = 102 \\ \Rightarrow \frac{r}{2} = (102 - 100) = 2 \\ \Rightarrow r = 4 \end{array}\]Putting \(r = 4\) in equation (i), we get:
\[\begin{array}{l} x \times 4 = 5000 \\ \Rightarrow x = \frac{5000}{4} = 1250 \end{array}\]Therefore, the sum of money is ₹ 1250 and the rate of interest is 4%.
Let the required sum of money = ₹ \(x\) and rate of interest = \(r\%\) per annum.
\[\begin{array}{l} \text{Simple interest for 2 years } = \frac{x \times r \times 2}{100} = \frac{xr}{50} \\ \text{As per question, } \frac{xr}{50} = 8400 \Rightarrow xr = 420000 \dots (i) \end{array}\] \[\begin{array}{l} \text{Compound interest for 2 years } = x \left\{ \left(1 + \frac{r}{100}\right)^2 - 1 \right\} \\ = x \left\{ 1 + \frac{2r}{100} + \frac{r^2}{10000} - 1 \right\} \\ = x \left( \frac{2r}{100} + \frac{r^2}{10000} \right) \\ = \frac{2xr}{100} + \frac{xr \cdot r}{10000} \end{array}\] \[\begin{array}{l} \text{As per question, } \frac{2xr}{100} + \frac{xr \cdot r}{10000} = 8652 \\ \Rightarrow \frac{2 \times 420000}{100} + \frac{420000 \times r}{10000} = 8652 \quad [\text{From (i)}] \\ \Rightarrow 8400 + 42r = 8652 \\ \Rightarrow 42r = (8652 - 8400) \\ \Rightarrow 42r = 252 \\ \Rightarrow r = \frac{252}{42} = 6 \end{array}\]Putting \(r = 6\) in equation (i), we get:
\[\begin{array}{l} x \times 6 = 420000 \\ \Rightarrow x = \frac{420000}{6} = 70000 \end{array}\]Therefore, the sum of money is ₹ 70000 and the rate of interest is 6%.
Principal \((P)\) = ₹ 6000
Rate of interest \((r)\) = 8% per annum
Time \((t)\) = 1 year = 2 half-years (since compounded at the interval of 6 months)
Rate of interest for 6 months = \(\frac{8}{2}\%\) = 4%
Therefore, amount at the end of 1 year
\[\begin{array}{l} = 6000 \times \left(1 + \frac{4}{100}\right)^2 \\ = 6000 \times \left(\frac{104}{100}\right)^2 \\ = 6000 \times \left(\frac{26}{25}\right)^2 \\ = 6000 \times \frac{676}{625} \\ = 9.6 \times 676 = 6489.60 \end{array}\]Therefore, Compound Interest = Amount - Principal
\[\begin{array}{l} = (6489.60 - 6000) \\ = 489.60 \end{array}\]Therefore, the required compound interest is ₹ 489.60.
Principal \((P)\) = ₹ 6250
Rate of interest \((r)\) = 10% per annum
Time \((t)\) = 9 months = \(\frac{9}{12}\) years = \(\frac{3}{4}\) years
Since interest is compounded at the interval of 3 months (quarterly), the number of conversion periods \((n) = \frac{9}{3} = 3\).
Amount \((A)\) after 9 months is calculated as:
\[\begin{array}{l} A = P \left(1 + \frac{r/4}{100}\right)^{4t} \\ = 6250 \left(1 + \frac{10/4}{100}\right)^3 \\ = 6250 \left(1 + \frac{10}{400}\right)^3 \\ = 6250 \left(1 + \frac{1}{40}\right)^3 \\ = 6250 \times \left(\frac{41}{40}\right)^3 \\ = 6250 \times \frac{68921}{64000} \\ = \frac{625 \times 68921}{6400} \\ = \frac{43075625}{6400} \\ = 6730.5664 \dots \end{array}\]Amount \(\approx\) ₹ 6730.57
Compound Interest = Amount - Principal
\[\begin{array}{l} = 6730.57 - 6250 \\ = 480.57 \end{array}\]Therefore, the required compound interest is ₹ 480.57.
Let the rate of compound interest be \(r\%\) per annum.
Principal \((P)\) = ₹ 60000
Amount \((A)\) = ₹ 69984
Time \((n)\) = 2 years
As per the formula for compound interest:
\[\begin{array}{l} A = P \left(1 + \frac{r}{100}\right)^n \\ 69984 = 60000 \left(1 + \frac{r}{100}\right)^2 \\ \Rightarrow \left(1 + \frac{r}{100}\right)^2 = \frac{69984}{60000} \\ \Rightarrow \left(1 + \frac{r}{100}\right)^2 = \frac{11664}{10000} \\ \Rightarrow \left(1 + \frac{r}{100}\right)^2 = \left(\frac{108}{100}\right)^2 \end{array}\]Taking square root on both sides:
\[\begin{array}{l} 1 + \frac{r}{100} = \frac{108}{100} \\ \Rightarrow \frac{r}{100} = \frac{108}{100} - 1 \\ \Rightarrow \frac{r}{100} = \frac{8}{100} \\ \Rightarrow r = 8 \end{array}\]Therefore, the rate of interest is 8% per annum.
Let the required time be \(n\) years.
Principal \((P)\) = ₹ 40000
Amount \((A)\) = ₹ 46656
Rate of interest \((r)\) = 8% per annum
As per the formula for compound interest:
\[\begin{array}{l} A = P \left(1 + \frac{r}{100}\right)^n \\ 46656 = 40000 \left(1 + \frac{8}{100}\right)^n \\ \Rightarrow \left(1 + \frac{8}{100}\right)^n = \frac{46656}{40000} \\ \Rightarrow \left(\frac{108}{100}\right)^n = \frac{11664}{10000} \\ \Rightarrow \left(\frac{108}{100}\right)^n = \left(\frac{108}{100}\right)^2 \end{array}\]Comparing the powers on both sides, we get:
\[\begin{array}{l} n = 2 \end{array}\]Therefore, the required time is 2 years.
Let the rate of compound interest be \(r\%\) per annum.
Principal \((P)\) = ₹ 10000
Amount \((A)\) = ₹ 12100
Time \((n)\) = 2 years
As per the formula for compound interest:
\[\begin{array}{l} A = P \left(1 + \frac{r}{100}\right)^n \\ 12100 = 10000 \left(1 + \frac{r}{100}\right)^2 \\ \Rightarrow \left(1 + \frac{r}{100}\right)^2 = \frac{12100}{10000} \\ \Rightarrow \left(1 + \frac{r}{100}\right)^2 = \frac{121}{100} \\ \Rightarrow \left(1 + \frac{r}{100}\right)^2 = \left(\frac{11}{10}\right)^2 \end{array}\]Taking square root on both sides:
\[\begin{array}{l} 1 + \frac{r}{100} = \frac{11}{10} \\ \Rightarrow \frac{r}{100} = \frac{11}{10} - 1 \\ \Rightarrow \frac{r}{100} = \frac{1}{10} \\ \Rightarrow r = \frac{100}{10} = 10 \end{array}\]Therefore, the rate of compound interest is 10% per annum.
Let the required time be \(n\) years.
Principal \((P)\) = ₹ 50000
Amount \((A)\) = ₹ 60500
Rate of interest \((r)\) = 10% per annum
As per the formula for compound interest:
\[\begin{array}{l} A = P \left(1 + \frac{r}{100}\right)^n \\ 60500 = 50000 \left(1 + \frac{10}{100}\right)^n \\ \Rightarrow \left(1 + \frac{1}{10}\right)^n = \frac{60500}{50000} \\ \Rightarrow \left(\frac{11}{10}\right)^n = \frac{605}{500} \\ \Rightarrow \left(\frac{11}{10}\right)^n = \frac{121}{100} \\ \Rightarrow \left(\frac{11}{10}\right)^n = \left(\frac{11}{10}\right)^2 \end{array}\]Comparing the powers on both sides, we get:
\[\begin{array}{l} n = 2 \end{array}\]Therefore, the required time is 2 years.
Let the required time be \(n\) years.
Principal \((P)\) = ₹ 300000
Amount \((A)\) = ₹ 399300
Rate of interest \((r)\) = 10% per annum
As per the formula for compound interest:
\[\begin{array}{l} A = P \left(1 + \frac{r}{100}\right)^n \\ 399300 = 300000 \left(1 + \frac{10}{100}\right)^n \\ \Rightarrow \left(1 + \frac{1}{10}\right)^n = \frac{399300}{300000} \\ \Rightarrow \left(\frac{11}{10}\right)^n = \frac{3993}{3000} \\ \Rightarrow \left(\frac{11}{10}\right)^n = \frac{1331}{1000} \\ \Rightarrow \left(\frac{11}{10}\right)^n = \left(\frac{11}{10}\right)^3 \end{array}\]Comparing the powers on both sides, we get:
\[\begin{array}{l} n = 3 \end{array}\]Therefore, the required time is 3 years.
Principal \((P)\) = ₹ 1600
Rate of interest \((r)\) = 10% per annum
Time \((t)\) = \(1\frac{1}{2}\) years = \(\frac{3}{2}\) years
Since the interest is compounded at the interval of 6 months (half-yearly), the number of conversion periods \((n) = \frac{3}{2} \times 2 = 3\).
Therefore, Amount \((A)\) after \(1\frac{1}{2}\) years is:
\[\begin{array}{l} A = P \left(1 + \frac{r/2}{100}\right)^{2t} \\ = 1600 \left(1 + \frac{10/2}{100}\right)^3 \\ = 1600 \left(1 + \frac{5}{100}\right)^3 \\ = 1600 \left(\frac{105}{100}\right)^3 \\ = 1600 \times \left(\frac{21}{20}\right)^3 \\ = 1600 \times \frac{9261}{8000} \\ = \frac{1600 \times 9261}{8000} \\ = \frac{9261}{5} \\ = 1852.20 \end{array}\]Amount = ₹ 1852.20
Compound Interest = Amount - Principal
\[\begin{array}{l} = 1852.20 - 1600 \\ = 252.20 \end{array}\]Therefore, the compound interest is ₹ 252.20 and the amount is ₹ 1852.20.
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